A 2.10 kg frictionless block is attached to an ideal spring with force constant 315N/m . Initially the block has velocity -3.75m/s and displacement 0.270m .

Find (a) the amplitude of the motion.
(b) the maximum acceleration of the block.
(c) the maximum force the spring exerts on the block.

Total mechanical energy

= (M/2)V^2 + (k/2)X^2 = constant
= 14.06 + 11.48 = 25.54 J
(using the given displacement and velocity data)

k = 315 N/m

(a) For the amplitude (Xmax), set V = 0 and solve
(k/2)X^2 = 25.54

for X

(b) maximum acceleration = k*Xmax/M
(c) k*Xmax

so is part a .162m

To find the amplitude of the motion, we can use the equation for the displacement of a mass-spring system:

x = A * cos(ωt + φ)

Where:
x is the displacement of the block (given as 0.270 m)
A is the amplitude of the motion (unknown)
ω is the angular frequency of the motion
t is the time
φ is the phase constant

Since the block starts from its equilibrium position, the initial phase angle is 0, which means φ = 0.

The equation for angular frequency, ω, is given by:

ω = √(k / m)

Where:
k is the force constant of the spring (given as 315 N/m)
m is the mass of the block (given as 2.10 kg)

Substituting the given values into the equation, we get:

ω = √(315 / 2.10) = √150 = 12.25 rad/s

Now, we can solve for the amplitude, A.

0.270 = A * cos(12.25 * 0 + 0)
0.270 = A * cos(0)
0.270 = A * 1

Therefore, the amplitude of the motion is A = 0.270 m.

To find the maximum acceleration of the block, we can use the equation:

amax = A * ω^2

Substituting the values we have:

amax = 0.270 * (12.25)^2 = 0.270 * 150.0625 = 40.51875 m/s^2

Therefore, the maximum acceleration of the block is 40.52 m/s^2.

To find the maximum force the spring exerts on the block, we can use Hooke's Law:

F = k * x

Substituting the values we have:

F = 315 * 0.270 = 85.05 N

Therefore, the maximum force the spring exerts on the block is 85.05 N.

To find the answers to these questions, we need to apply the principles of simple harmonic motion (SHM) and Hooke's Law.

(a) To find the amplitude of the motion:
In simple harmonic motion, the displacement of an object is given by the equation:
x(t) = A * cos(ωt + φ)

Where:
x(t) is the displacement at time t,
A is the amplitude of motion,
ω is the angular frequency,
t is time, and
φ is the phase constant.

From the given information, we know that the initial displacement (x) is 0.270 m. At this point, the object is at its maximum displacement from the equilibrium position. Therefore, we can set the displacement equation equal to the given displacement value and solve for the amplitude (A):

0.270 m = A * cos(φ)

Since the block is initially at its maximum displacement, the cosine term is at its maximum value of 1. Therefore:

A = 0.270 m

So, the amplitude of motion is 0.270 m.

(b) To find the maximum acceleration of the block:
In simple harmonic motion, the acceleration of an object is given by:

a(t) = -ω^2 * x(t)

Where:
a(t) is the acceleration at time t,
ω is the angular frequency,
x(t) is the displacement at time t.

We can find the angular frequency (ω) using the equation:

ω = sqrt(k / m)

Where:
k is the force constant of the spring (315 N/m), and
m is the mass of the block (2.10 kg).

Using the given values, we can calculate ω:

ω = sqrt(315 N/m / 2.10 kg) = 12.19 rad/s

Now, we can substitute the angular frequency into the acceleration equation:

a(t) = -ω^2 * x(t) = -(12.19 rad/s)^2 * 0.270 m = -99.33 m/s^2

So, the maximum acceleration of the block is 99.33 m/s^2.

(c) To find the maximum force the spring exerts on the block:
According to Hooke's Law, the force exerted by a spring is given by:

F = -k * x(t)

Where:
F is the force exerted by the spring,
k is the force constant of the spring, and
x(t) is the displacement of the block at time t.

Using the given force constant (k = 315 N/m) and maximum displacement (x = 0.270 m), we can calculate the maximum force:

F = -k * x(t) = -(315 N/m) * (0.270 m) = -85.05 N

Since the force is negative, it means that the maximum force the spring exerts on the block is 85.05 N in the opposite direction of the displacement.

So, the answers are:
(a) The amplitude of the motion is 0.270 m.
(b) The maximum acceleration of the block is 99.33 m/s^2.
(c) The maximum force the spring exerts on the block is 85.05 N.